Question

write the set builder form of the set A={1,2,5,12,27.....}​

Answer 1

Consider the terms in the set A,

1, 2, 5, 12, 27,...

The terms are formed as follows:

1 × 2 + 0 = 2

2 × 2 + 1 = 5

5 × 2 + 2 = 12

12 × 2 + 3 = 27

...

So the 2nd term 2 is obtained as,

$\longrightarrow\sf{2=0+2\times1}$

$\longrightarrow\sf{2=0\times2^0+2^1}$

And 3rd term 5 is obtained as,

$\longrightarrow\sf{5=1+2\times2}$

$\longrightarrow\sf{5=1+2(0+2)}$

$\longrightarrow\sf{5=1\times2^0+0\times2^1+2^2}$

And 4th term 12 is obtained as,

$\longrightarrow\sf{12=2+2\times5}$

$\longrightarrow\sf{12=2+2(1+2^2)}$

$\longrightarrow\sf{12=2\times2^0+1\times2^1+0\times2^2+2^3}$

And 5th term 27 is obtained as,

$\longrightarrow\sf{27=3+2\times12}$

$\longrightarrow\sf{27=3+2(2+2+2^3)}$

$\longrightarrow\sf{27=3\times2^0+2\times2^1+1\times2^2+0\times2^3+2^4}$

So the 6th term will be,

$\longrightarrow\sf{a_6=4\times2^0+3\times2^1+2\times2^2+1\times2^3+0\times2^4+2^5}$

$\longrightarrow\sf{a_6=4\times1+3\times2+2\times4+1\times8+0\times16+32}$

$\longrightarrow\sf{a_6=4+6+8+8+32}$

$\longrightarrow\sf{a_6=58}$

Thus the $\sf{n^{th}}$ term of the sequence is,

$\longrightarrow\sf{a_n=(n-2)2^0+(n-3)2^1+\,\dots\,+1\times2^{n-3}+0\times2^{n-2}+2^{n-1}}$

$\longrightarrow\sf{a_n=\displaystyle2^{n-1}+\sum_{r=0}^{n-2}(n-2-r)\cdot2^r}$

Hence the set A in set builder form is,

$\longrightarrow\large\text{ \displaystyle\mathsf{\underline{\underline{A=\left\{x:x=2^{n-1}+\sum_{r=0}^{n-2}(n-2-r)\cdot2^r,\ n\in\mathbb{N}\right\}}}} }$